Mathematics, 30.11.2019 01:31 bettybales1986

Arandom sample of 378 hotel guests was taken one year ago, and it was found that 178 requested nonsmoking rooms. recently, a random sample of 516 hotel guests showed that 320 requested nonsmoking rooms. do these data indicate that the proportion of hotel guests requesting nonsmoking rooms has increased?

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The p value is a very low value and using any significance level for example $\alpha=0.05, 0,1,0.15$ always p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion of proportion of hotel guests requesting nonsmoking rooms has increased.

Step-by-step explanation:

1) Data given and notation

$X_{B}=178$ represent the number of requested nonsmoking rooms before

$X_{A}=320$ represent the number of requested nonsmoking rooms after

$n_{B}=378$ sample of number of requested nonsmoking rooms before

$n_{A}=516$ sample of number of requested nonsmoking rooms after

$p_{B}=\frac{178}{378}=0.471$ represent the proportion of requested nonsmoking rooms before

$p_{A}=\frac{320}{516}=0.620$ represent the proportion of requested nonsmoking rooms after

z would represent the statistic (variable of interest)

p_v represent the value for the test (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the proportion of requested nonsmoking rooms after is higher then the proportion before , the system of hypothesis would be:

Null hypothesis: $p_{A} \leq p_{B}$

Alternative hypothesis: $p_{A} p_{B}$

We need to apply a z test to compare proportions, and the statistic is given by:

$z=\frac{p_{A}-p_{B}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{A}}+\frac{1}{n_{B}})}}$ (1)

Where $\hat p=\frac{X_{A}+X_{B}}{n_{A}+n_{B}}}=\frac{178+320}{378+516}=0.557$

3) Calculate the statistic

Replacing in formula (1) the values obtained we got this:

$z=\frac{0.620-0.471}{\sqrt{0.557(1-0.557)(\frac{1}{516}+\frac{1}{378})}}=4.43$

4) Statistical decision

For this case we don't have a significance level provided $\alpha$ , but we can calculate the p value for this test.

Since is a one side test the p value would be:

$p_v =P(Z4.43)=4.71x10^{-6}$

So the p value is a very low value and using any significance level for example $\alpha=0.05, 0,1,0.15$ always p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion of proportion of hotel guests requesting nonsmoking rooms has increased.

Answer from: Quest

step-by-step explanation: so we look at the formula un = a + (n-1)d

un is the amount of seats

n is the nth term aka row number

a is the original sequence/row number

d is the difference between each term

so in this a=22 d=6 and n can remain as n because we are simply formjng a rule

lets put these in the equation

un=22+(n-1)(6)

this is your answer to part a

however

lets double check that this is correct, we'll try putting it into the second term

un=a+(n-1)d

u2=22+(2-1)(6)

u2=22+6

u2=28

as the question said the second row has 28 seats so now we know the formula we made is correct.

as for part b

un=a+(n-1)

its asking us to find a row with 100 seats.

remember un was the amount of seata so all we do is substitute this in.

un=100 a=22 d=6

un=a+(n-1)d

100=22+(n-1)(6)

so now we simplify to find n which is the row number.

100-22=22+(n-1)(6)-22

78=(n-1)(6)

78/6=(n-1)(6)/6

13=n-1

13+1=n-1+1

14=n

so now we know that in the 14th row there is 100 seats.

hope this and that you understood! its my first time using the app or anyone out on here : p its also me to revise so a bunch! ask more questions if you dont understand ☺

Answer from: Quest

$17x-13$